Calculating VA, load, for UPS

[vicw]

Hi,
I am looking for a UPS. I want to calculate VA.
I know the max watts my computer draws. I know that I would have to multiply this value by 1.4 to obtain the VA, load. Or do I use the max power rating on my power supply to obtain the VA, load?
Thanks,
Vic

 

[daKlone]

If you know the max watts the unit requires, then you know the VA load (Watts = V*A = VA).

When sizing a UPS, it very much depends on how you want your UPS to perform, and whether you're going to have a monitor (about 100W), printer (about 15W) etc. attached to it as well. The bigger the load, the larger (and more expensive) UPS you need. Although of course, for the same load a larger UPS will give you more endurance (it will hold up your system for longer).

You can either measure the AC load current on each of your units in various working conditions (hard/floppy drives running, printer printing etc.) to get an idea of the max load, or it's probably easier to take the max power rating from the PSU plates of the items you want to power up and add them together.

Hope this helps.

 

[vicw]

daklone,

On the APC, American Power Conversion, website, they allow the user to select pc components. They show a value in Watts and then they give a value in VA. I think it has something to do with the power factor... i'm not too sure what it is though. APC says if the value you have is in Watts multiply that value by 1.4.
For my power supply it can dish out 300W. It converts AC to DC, since the 300W is DC only I believe that to calculate the input AC they multiply by a factor of 1.4.

Any opinions?

 

[daKlone]

I see. Perhaps we got a bit confused by the x1.4 thing.

Strictly, Watts = V*A*Pf. The trouble being that (especially) with consumer equipment, you don't know what the power factor (Pf) is. You are then faced with the prospect of trying to find it out (from the manufacturer or by measurement) or making an assumption. The assumption you make depends on what sort of equipment it is (inductive loads are bad for Pf) and if it has any kind of PFC (power factor correction) circuitry. Ideally, Pf = 1 (Watts = V*A*1) but it can vary widely.

But that has nothing to do with your x1.4.

When APC are telling you to multiply DC power by 1.4, they are in fact trying to make a (probably fair) assumption that your AC/DC PSU is about 60% efficient (i.e. it 'wastes' 40% of the power it consumes as heat etc.).

This assumption probably errs on the side of caution (assumes that your PSU is fully-loaded) which might give you a larger UPS than you actually need (but larger = longer endurance = good). My rule-of-thumb is, as I said, to use the power rating on the equipment plate. Often a PSU has input AC Amps/Volts written on it.

Just as a note, on the APC website there is a UPS sizer that apparently uses measured data from a variety of equipment to suggest a UPS. They also have an Express Selector version that allows you to enter power requirements directly in Watts, VA, kVA or Amps. Give it a try and see what it comes up with.

Hope this clears things up a bit.

 

[vicw]

On my PSU it shows an input of 120V @ 7A = 840W. This value is way above the output power I measured from the PSU, 244W. This value is way below... I did use the APC selector and it gave me a value similar to mine but with the x1.4.
I know that for my PSU there is a power factor correction (don't know what that is). It states: the power module shall meet the IEC- 1000-3-2 Standard.
Since I know how much real power I am using can I use my value rather than the value from the PSU plate? I know it may be safer to have a larger=longer=good= but more expensive.
I'm not too sure about this, but I don't think my power supply will ever draw 840W from the wall.
daKlone, I understand your point, but I guess what I'm really looking for is the power factor of the supply.
Thanks for your help.

 

[daKlone]

I agree, 840W does seem a bit extortionate! I don't think I've EVER seen a PC consume that amount of power! Normally it's more like 500W max!

How critical is your application? Do you have a closely-defined hold-up time? (does it matter if the hold-up is, say, 20 minutes instead of 25 or vice-versa?).

If you unit states that it has power factor correction, then it's Pf should be at least 0.97.

Provided you are happy with your 244W measurement, and it's not a time-critical application then I would go with a 60% PSU effiecency giving a load of about 342W (not including monitor etc).

Anyway, why not measure the AC input current while your using the machine? Put it through it's paces and you'll soon get a good idea of how much power it's really using.